\(\int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx\) [193]

Optimal result

Integrand size = 16, antiderivative size = 116 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=-\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^2}{2 b^4}+\frac {B x^3}{3 b^3}-\frac {a^4 (A b-a B)}{2 b^6 (a+b x)^2}+\frac {a^3 (4 A b-5 a B)}{b^6 (a+b x)}+\frac {2 a^2 (3 A b-5 a B) \log (a+b x)}{b^6} \]

[Out]

-3*a*(A*b-2*B*a)*x/b^5+1/2*(A*b-3*B*a)*x^2/b^4+1/3*B*x^3/b^3-1/2*a^4*(A*b-B*a)/b^6/(b*x+a)^2+a^3*(4*A*b-5*B*a)
/b^6/(b*x+a)+2*a^2*(3*A*b-5*B*a)*ln(b*x+a)/b^6

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=-\frac {a^4 (A b-a B)}{2 b^6 (a+b x)^2}+\frac {a^3 (4 A b-5 a B)}{b^6 (a+b x)}+\frac {2 a^2 (3 A b-5 a B) \log (a+b x)}{b^6}-\frac {3 a x (A b-2 a B)}{b^5}+\frac {x^2 (A b-3 a B)}{2 b^4}+\frac {B x^3}{3 b^3} \]

[In]

Int[(x^4*(A + B*x))/(a + b*x)^3,x]

[Out]

(-3*a*(A*b - 2*a*B)*x)/b^5 + ((A*b - 3*a*B)*x^2)/(2*b^4) + (B*x^3)/(3*b^3) - (a^4*(A*b - a*B))/(2*b^6*(a + b*x
)^2) + (a^3*(4*A*b - 5*a*B))/(b^6*(a + b*x)) + (2*a^2*(3*A*b - 5*a*B)*Log[a + b*x])/b^6

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 a (-A b+2 a B)}{b^5}+\frac {(A b-3 a B) x}{b^4}+\frac {B x^2}{b^3}-\frac {a^4 (-A b+a B)}{b^5 (a+b x)^3}+\frac {a^3 (-4 A b+5 a B)}{b^5 (a+b x)^2}-\frac {2 a^2 (-3 A b+5 a B)}{b^5 (a+b x)}\right ) \, dx \\ & = -\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^2}{2 b^4}+\frac {B x^3}{3 b^3}-\frac {a^4 (A b-a B)}{2 b^6 (a+b x)^2}+\frac {a^3 (4 A b-5 a B)}{b^6 (a+b x)}+\frac {2 a^2 (3 A b-5 a B) \log (a+b x)}{b^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=\frac {18 a b (-A b+2 a B) x+3 b^2 (A b-3 a B) x^2+2 b^3 B x^3+\frac {3 a^4 (-A b+a B)}{(a+b x)^2}+\frac {6 a^3 (4 A b-5 a B)}{a+b x}-12 a^2 (-3 A b+5 a B) \log (a+b x)}{6 b^6} \]

[In]

Integrate[(x^4*(A + B*x))/(a + b*x)^3,x]

[Out]

(18*a*b*(-(A*b) + 2*a*B)*x + 3*b^2*(A*b - 3*a*B)*x^2 + 2*b^3*B*x^3 + (3*a^4*(-(A*b) + a*B))/(a + b*x)^2 + (6*a
^3*(4*A*b - 5*a*B))/(a + b*x) - 12*a^2*(-3*A*b + 5*a*B)*Log[a + b*x])/(6*b^6)

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {-\frac {1}{3} b^{2} B \,x^{3}-\frac {1}{2} A \,b^{2} x^{2}+\frac {3}{2} B a b \,x^{2}+3 a A b x -6 a^{2} B x}{b^{5}}+\frac {2 a^{2} \left (3 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}-\frac {a^{4} \left (A b -B a \right )}{2 b^{6} \left (b x +a \right )^{2}}+\frac {a^{3} \left (4 A b -5 B a \right )}{b^{6} \left (b x +a \right )}\)	\(116\)
norman	\(\frac {\frac {a^{2} \left (9 a^{2} b A -15 a^{3} B \right )}{b^{6}}+\frac {B \,x^{5}}{3 b}+\frac {\left (3 A b -5 B a \right ) x^{4}}{6 b^{2}}-\frac {2 a \left (3 A b -5 B a \right ) x^{3}}{3 b^{3}}+\frac {2 a \left (6 a^{2} b A -10 a^{3} B \right ) x}{b^{5}}}{\left (b x +a \right )^{2}}+\frac {2 a^{2} \left (3 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}\)	\(120\)
risch	\(\frac {B \,x^{3}}{3 b^{3}}+\frac {A \,x^{2}}{2 b^{3}}-\frac {3 B a \,x^{2}}{2 b^{4}}-\frac {3 a A x}{b^{4}}+\frac {6 a^{2} B x}{b^{5}}+\frac {\left (4 A \,a^{3} b -5 B \,a^{4}\right ) x +\frac {a^{4} \left (7 A b -9 B a \right )}{2 b}}{b^{5} \left (b x +a \right )^{2}}+\frac {6 a^{2} \ln \left (b x +a \right ) A}{b^{5}}-\frac {10 a^{3} \ln \left (b x +a \right ) B}{b^{6}}\)	\(123\)
parallelrisch	\(\frac {2 b^{5} B \,x^{5}+3 A \,b^{5} x^{4}-5 B a \,b^{4} x^{4}+36 A \ln \left (b x +a \right ) x^{2} a^{2} b^{3}-12 A a \,b^{4} x^{3}-60 B \ln \left (b x +a \right ) x^{2} a^{3} b^{2}+20 B \,a^{2} b^{3} x^{3}+72 A \ln \left (b x +a \right ) x \,a^{3} b^{2}-120 B \ln \left (b x +a \right ) x \,a^{4} b +36 A \ln \left (b x +a \right ) a^{4} b +72 a^{3} b^{2} A x -60 B \ln \left (b x +a \right ) a^{5}-120 a^{4} b B x +54 a^{4} b A -90 a^{5} B}{6 b^{6} \left (b x +a \right )^{2}}\)	\(186\)

[In]

int(x^4*(B*x+A)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/b^5*(-1/3*b^2*B*x^3-1/2*A*b^2*x^2+3/2*B*a*b*x^2+3*a*A*b*x-6*a^2*B*x)+2*a^2*(3*A*b-5*B*a)*ln(b*x+a)/b^6-1/2*
a^4*(A*b-B*a)/b^6/(b*x+a)^2+a^3*(4*A*b-5*B*a)/b^6/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.70 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=\frac {2 \, B b^{5} x^{5} - 27 \, B a^{5} + 21 \, A a^{4} b - {\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{3} + 3 \, {\left (21 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{2} + 6 \, {\left (B a^{4} b + A a^{3} b^{2}\right )} x - 12 \, {\left (5 \, B a^{5} - 3 \, A a^{4} b + {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^3,x, algorithm="fricas")

[Out]

1/6*(2*B*b^5*x^5 - 27*B*a^5 + 21*A*a^4*b - (5*B*a*b^4 - 3*A*b^5)*x^4 + 4*(5*B*a^2*b^3 - 3*A*a*b^4)*x^3 + 3*(21
*B*a^3*b^2 - 11*A*a^2*b^3)*x^2 + 6*(B*a^4*b + A*a^3*b^2)*x - 12*(5*B*a^5 - 3*A*a^4*b + (5*B*a^3*b^2 - 3*A*a^2*
b^3)*x^2 + 2*(5*B*a^4*b - 3*A*a^3*b^2)*x)*log(b*x + a))/(b^8*x^2 + 2*a*b^7*x + a^2*b^6)

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.17 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=\frac {B x^{3}}{3 b^{3}} - \frac {2 a^{2} \left (- 3 A b + 5 B a\right ) \log {\left (a + b x \right )}}{b^{6}} + x^{2} \left (\frac {A}{2 b^{3}} - \frac {3 B a}{2 b^{4}}\right ) + x \left (- \frac {3 A a}{b^{4}} + \frac {6 B a^{2}}{b^{5}}\right ) + \frac {7 A a^{4} b - 9 B a^{5} + x \left (8 A a^{3} b^{2} - 10 B a^{4} b\right )}{2 a^{2} b^{6} + 4 a b^{7} x + 2 b^{8} x^{2}} \]

[In]

integrate(x**4*(B*x+A)/(b*x+a)**3,x)

[Out]

B*x**3/(3*b**3) - 2*a**2*(-3*A*b + 5*B*a)*log(a + b*x)/b**6 + x**2*(A/(2*b**3) - 3*B*a/(2*b**4)) + x*(-3*A*a/b
**4 + 6*B*a**2/b**5) + (7*A*a**4*b - 9*B*a**5 + x*(8*A*a**3*b**2 - 10*B*a**4*b))/(2*a**2*b**6 + 4*a*b**7*x + 2
*b**8*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=-\frac {9 \, B a^{5} - 7 \, A a^{4} b + 2 \, {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x}{2 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} + \frac {2 \, B b^{2} x^{3} - 3 \, {\left (3 \, B a b - A b^{2}\right )} x^{2} + 18 \, {\left (2 \, B a^{2} - A a b\right )} x}{6 \, b^{5}} - \frac {2 \, {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x + a\right )}{b^{6}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(9*B*a^5 - 7*A*a^4*b + 2*(5*B*a^4*b - 4*A*a^3*b^2)*x)/(b^8*x^2 + 2*a*b^7*x + a^2*b^6) + 1/6*(2*B*b^2*x^3
- 3*(3*B*a*b - A*b^2)*x^2 + 18*(2*B*a^2 - A*a*b)*x)/b^5 - 2*(5*B*a^3 - 3*A*a^2*b)*log(b*x + a)/b^6

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=-\frac {2 \, {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6}} - \frac {9 \, B a^{5} - 7 \, A a^{4} b + 2 \, {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{6}} + \frac {2 \, B b^{6} x^{3} - 9 \, B a b^{5} x^{2} + 3 \, A b^{6} x^{2} + 36 \, B a^{2} b^{4} x - 18 \, A a b^{5} x}{6 \, b^{9}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^3,x, algorithm="giac")

[Out]

-2*(5*B*a^3 - 3*A*a^2*b)*log(abs(b*x + a))/b^6 - 1/2*(9*B*a^5 - 7*A*a^4*b + 2*(5*B*a^4*b - 4*A*a^3*b^2)*x)/((b
*x + a)^2*b^6) + 1/6*(2*B*b^6*x^3 - 9*B*a*b^5*x^2 + 3*A*b^6*x^2 + 36*B*a^2*b^4*x - 18*A*a*b^5*x)/b^9

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.27 \[ \int \frac {x^4 (A+B x)}{(a+b x)^3} \, dx=x^2\,\left (\frac {A}{2\,b^3}-\frac {3\,B\,a}{2\,b^4}\right )-x\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{b}+\frac {3\,B\,a^2}{b^5}\right )-\frac {x\,\left (5\,B\,a^4-4\,A\,a^3\,b\right )+\frac {9\,B\,a^5-7\,A\,a^4\,b}{2\,b}}{a^2\,b^5+2\,a\,b^6\,x+b^7\,x^2}-\frac {\ln \left (a+b\,x\right )\,\left (10\,B\,a^3-6\,A\,a^2\,b\right )}{b^6}+\frac {B\,x^3}{3\,b^3} \]

[In]

int((x^4*(A + B*x))/(a + b*x)^3,x)

[Out]

x^2*(A/(2*b^3) - (3*B*a)/(2*b^4)) - x*((3*a*(A/b^3 - (3*B*a)/b^4))/b + (3*B*a^2)/b^5) - (x*(5*B*a^4 - 4*A*a^3*
b) + (9*B*a^5 - 7*A*a^4*b)/(2*b))/(a^2*b^5 + b^7*x^2 + 2*a*b^6*x) - (log(a + b*x)*(10*B*a^3 - 6*A*a^2*b))/b^6
+ (B*x^3)/(3*b^3)